\documentclass[11pt]{article}
\usepackage{geometry}                % See geometry.pdf to learn the layout options. There are lots.
\geometry{letterpaper}                   % ... or a4paper or a5paper or ... 
%\geometry{landscape}                % Activate for for rotated page geometry
%\usepackage[parfill]{parskip}    % Activate to begin paragraphs with an empty line rather than an indent
\usepackage{graphicx}
\usepackage{amssymb,amsmath}
\usepackage{epstopdf}
\DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png}

\title{Expanding the GW signal explicitly}
\author{Anand Sengupta}
%\date{}                                           % Activate to display a given date or no date

\begin{document}
\maketitle
%\section{}
%\subsection{}

\section{Re-parametrization of the inspiral signal}

As defined in JKS98, the antennae pattern functions are given by 
\begin{eqnarray} 
\label{eq:FpFc} 
F_+ &=& A \cos(2\psi) + B \sin(2\psi), \\ F_+ &=& -A \sin(2\psi) + B \cos(2\psi), 
\end{eqnarray} 
where, $\psi$ is the so called polarization angle of the source 
and $A(\alpha, \delta)$ and $B(\alpha, \delta)$ are functions of the sky positions only. 
$\alpha$ is the right-ascension of the source and $\delta$ is the declination. These 
themselves are function of time due to the Earth's motion. However, we will consider them 
to be constant over the period of the inspiral (typically few tens of seconds). Further 
calculating them at time when $\rm{GMSTRad}=0$ lines it up nicely.

The most general form of the inspiral signal can be written as 
\begin{equation} 
\label{eq:signal} h(t) = \frac{2M\eta}{r} x \{ F_{+} h_{+} + F_{\times} h_{\times} \}, 
\end{equation} 
where, 
\begin{equation} x \equiv (m \omega)^{2/3}, 
\end{equation} 
and, 
\begin{eqnarray} 
h_{+}	&=& -(1 + \cos^2 \iota) \cos(\Phi + \Phi_0), \\ 
h_{\times}	&=& -2 \cos \iota \sin(\Phi + \Phi_0). 
\end{eqnarray} 
Here, $\omega$ is the system orbital 
frequency, $M$ and $\eta$ are the total mass and symmetric mass ratio respectively and $r$ 
is the distance to the system from the centre of the Earth. Also, in the above set of 
equations, $\iota$ is the inclination angle (the angle between the binary's orbital plane 
to the line of sight from the centre of the Earth), $\Phi$ is the 3.5PN phase and $\Phi_0$ 
is some offset in the phase.

Putting Equations (1,2) and (5,6) in Equation 3, we can show that the most general form of 
the binary inspiral signal (in the restricted PN approximation) can be written as a sum of 
4 terms: 
\begin{equation} 
h(t) = \sum_{k=1}^4 a_k \; h_k(t) 
\end{equation} 
where, 
\begin{eqnarray} 
a_1 &=& \frac{1}{r} \left [ -(1 + \cos^2 \iota) \cos \Phi_0 \sin(2\psi) -  2 \cos \iota \cos(2\psi) \sin \Phi_0 \right ], \\ 
a_2 &=& \frac{1}{r} \left [ (1 + \cos^2 \iota) \sin \Phi_0 \cos(2\psi) + 2 \cos \iota \cos(2\psi) \sin \Phi_0 \right ], \\ 
a_3 &=& \frac{1}{r} \left [ -(1 + \cos^2 \iota) \cos \Phi_0 \cos(2\psi) + 2 \cos \iota \sin(2\psi) \sin \Phi_0 \right ], \\ 
a_4 &=& \frac{1}{r} \left [ (1 + \cos^2 \iota) \sin \Phi_0 \sin(2\psi) - 2 \cos \iota \cos(2\psi) \cos \Phi_0 \right ], 
\end{eqnarray} and 
\begin{eqnarray} 
h_1 &=& (2 M \eta)\; B \; x \; \cos\Phi, \\ 
h_2 &=& (2 M \eta)\; A\; x \;  \sin\Phi, \\ 
h_3 &=& (2 M \eta)\; A\; x \; \cos\Phi, 
\\ h_4 &=& (2 M \eta)\; B\; x \; \sin\Phi . 
\end{eqnarray} 
It is easy to see that the set $\{a_k\}$ are functions of the 
extrinsic parameters $(r, \Phi_0, \iota, \psi)$. This re-parametrization makes the 
expression linear in $a_k$.


% \exp \left[ (a_1 B + a_3 A) - {\rm{i}}(a_4 B + a_2 A) \right ]
\section{SPA Fourier transform of the signal}

It can be shown that in the Fourier domain, the signal can be written as
\begin{equation} 
\tilde{h}(f) = A_{\rm{eff}} \; f^{-7/6} \; \exp \left [ {\mathbf{i}} \; \Psi(f; M, \eta, t_c) \right ]\;
\end{equation} 
where, 
\begin{equation} 
A_{\rm{eff}} = \sqrt{\frac{5\eta}{96}}  M^{5/6} \pi^{-2/3} \; \left [ (a_1 B + a_3 A) - {\mathbf{i}}(a_4 B + a_2 A) \right ] . 
\end{equation} 
Note that ${\mathbf{i}}^2 = -1$ and the phase $\Psi$ is given by 
\begin{equation} 
\Psi(f; M, \eta, t_c) = 2\pi f t_c + \Phi - \frac{\pi}{4}. 
\end{equation}

It is easy to see that the signal power spectral density $| \tilde{h}(f) | ^2$ is given by 
\begin{equation}
| \tilde{h}(f) | ^2 = \frac{5\eta}{96} M^{5/3} \pi^{-4/3} \; \left[(a_1 B + a_3 A)^2 + (a_4 B + a_2 A)^2 \right] \; 	f^{-7/3}
\end{equation}


\subsection{Averaging over angles} Sometimes, it will be convenient to express the waveform 
by averaging over the inclination angle $\iota$, polarization angle $\psi$ and phase offset 
$\Phi_0$.

The RMS averaged Fourier domain signal can be expressed as, 
\begin{equation} 
\left < \tilde{h}(f) \right >_{\rm{angles}} = \left < A_{\rm{eff}}\right >_{\rm{angles}} \; f^{-7/6} \; \exp \left [ {\mathbf{i}} \; \Psi(f; M, \eta, t_c) \right ]\; 
\end{equation} 
where, 
\begin{equation} \left < A_{\rm{eff}}\right >_{\rm{angles}} = \sqrt{\frac{5\eta}{96}} M^{5/6} \pi^{-2/3} \; \left [ \left < a_1 B + a_3 A \right >_{\rm{angles}} - {\mathbf{i}} \left < a_4 B + a_2 A\right >_{\rm{angles}} \right ] 
\end{equation}

The RMS averages on the right hand side of the above equation can be explicitly computed: 
\begin{eqnarray}
 \left <  a_1 B + a_3 A \right >_{\iota, \Phi_0, \psi} &=&  \sqrt{ \frac{1}{\pi^3} \int_0^{\pi/2} \; d\iota \int_0^{\pi} \; d\psi \int_{-\pi}^{\pi} \; d\Phi_0 \; (a_1 B + a_3 A)^2},\\
 									&=& \frac{1}{r} 
\sqrt{\frac{35 (A^2 + B^2)}{32}}. \end{eqnarray} Similarly, \begin{eqnarray}
 \left <  a_1 B + a_3 A \right >_{\Phi_0, \psi} &=&  \sqrt{ \frac{1}{2 \pi^2} \int_0^{\pi} \; d\psi \int_{-\pi}^{\pi} \; d\Phi_0 \; (a_1 B + a_3 A)^2},\\
 							&=& \frac{1}{r} \sqrt{	\frac{35 + 28 \cos(2\iota) + \cos(4\iota)}{32}	}
\end{eqnarray}

The angle averaged signal in the Fourier domain can then be written as
\begin{equation}
\left < \tilde{h}(f) \right >_{\iota, \Phi_0, \psi} =  \sqrt{\frac{35(A^2 + B^2)}{32}} \sqrt{\frac{5\eta}{96}}\; \frac{M^{5/6} \pi^{-2/3}}{r}\; (1 - {\mathbf{i}}) \; f^{-7/6} \; \exp \left [ {\mathbf{i}} \; \Psi(f; M, \eta, t_c) \right ]\;
\end{equation}


\section{General expression for the Fisher matrix elements}

As seen above, the SPA Fourier transformed signal has the following generic form
\begin{equation}
\tilde{h}(f) =  \left( P + {\mathbf{i}}\; Q \right) f^{-7/6} \exp\left [ {{\mathbf{i}} \; \Psi} \right ],
\end{equation}
where $P, Q$ and $\Psi$ are functions of the parameters $\{x_\alpha\}$. They can be explicitly written as,
\begin{eqnarray}
P &=& \sqrt{\frac{5\eta}{96}} M^{5/6} \pi^{-2/3} (a_1 B + a_3 A) \\
Q &=& -\sqrt{\frac{5\eta}{96}} M^{5/6} \pi^{-2/3} (a_4 B + a_2 A).
\end{eqnarray}

For the case of waveform averaged over the angles $(\iota, \psi, \Phi_0)$, the form of $P$ and $Q$ become
\begin{eqnarray}
P &=& \sqrt{\frac{35(A^2 + B^2)}{32}} \sqrt{\frac{5\eta}{96}}\; \frac{M^{5/6} \pi^{-2/3}}{r} \\
Q &=& -P
\end{eqnarray}

The Fisher matrix element 
${\mathcal{F}}_{\alpha\beta}$ can be expressed in terms of the derivatives of $P, Q$ and $\Psi$ as shown below:
\begin{eqnarray}
{\mathcal{F}}_{\alpha\beta} &=& \left < \tilde{h}_{,\alpha} | \tilde{h}_{,\beta} \right > \\
					&=& 4 \int_0^{f_{\rm{LSO}}} \frac{\tilde{h}_{,\alpha} \tilde{h}_{,\beta}^* }{S_h(f)} \; df \\
					&=& 4 \int_0^{f_{\rm{LSO}}} \frac{( P_{,\alpha }  P_{,\beta }  + Q _{,\alpha }  Q_{,\beta }) + (P^2 + Q^2 )\Psi_{,\alpha }  \Psi_{,\beta }}{S_h(f)} f^{-7/3} \; df
\end{eqnarray}
This is useful from the point of view of coding.


\end{document}  
